Let $x_t = O_p(1)$, meaning that for all $\varepsilon > 0$ there exists $M_{\varepsilon} < \infty$ s.t. $P(|X_t| > M_{\varepsilon}) < \epsilon$ for all $t \in \mathbb{N}$. Does it imply that $\sup_{t} Ex_t < \infty$, namely, whether the expectation of a random variable which is bounded in probability is finite. If it is not the case, could you give me a counter example?

  • $\begingroup$ Could you please explain your notation? It could mean practically anything... . $\endgroup$ – whuber Jul 15 '13 at 18:38
  • $\begingroup$ I tried to make it more understandable. I don't really understand where is a problem, since I used a standard notation, I am sorry about it. $\endgroup$ – Kolibris Jul 15 '13 at 18:48
  • $\begingroup$ Thanks. The appearance of $p$ and $t$ as subscripts is mystifying: do these have any meaning? $\endgroup$ – whuber Jul 15 '13 at 18:48
  • $\begingroup$ A subscript $p$ is for bounded in PROBABILITY, $t$ is just an index, since I am considering time series, but it is not very important. I made it explicit. Thank you very much for your comments. $\endgroup$ – Kolibris Jul 15 '13 at 18:54
  • $\begingroup$ I believe $t$ is essential, for otherwise it looks like you are merely asking for an example of a random variable whose expectation is not defined. Indeed, it is difficult to see how your question differs substantially from that. With the introduction of the index $t$, it is crucial that you supply a quantifier in your statement: do you mean there exists a $t$ for which $x_t$ has finite expectation or for all $t$ the expectation should be finite, or perhaps for infinitely many or almost all $t$? $\endgroup$ – whuber Jul 15 '13 at 19:00

Here is a counter example. Let $X_{n}$, $n=1,2,\ldots$, be a sequence of random variables, whose distributions are defined as follows, $$ X_{n}=\begin{cases} 0, & \mbox{with probability }\frac{n-1}{n},\\ n^{2}, & \mbox{with probability }\frac{1}{n}. \end{cases} $$ To see $X_{n}\sim O_{p}\left(1\right)$, let $\varepsilon>0$ be an arbitrary positive number, and there exists $M_{\varepsilon}=\left\lfloor \varepsilon^{-1}+1\right\rfloor ^{2}$ ($\left\lfloor \varepsilon^{-1}\right\rfloor $ is the integer part of $\varepsilon^{-1}$) such that \begin{eqnarray*} \sup_{n}\Pr\left(\left|X_{n}\right|\geq M_{\varepsilon}\right) & = & \Pr\left(X_{\left\lfloor \varepsilon^{-1}+1\right\rfloor }=\left(\left\lfloor \varepsilon^{-1}+1\right\rfloor \right)^{2}\right)\\ & = & \frac{1}{\left\lfloor \varepsilon^{-1}+1\right\rfloor }\\ & < & \varepsilon. \end{eqnarray*} However, $\mathrm{{E}}\left(X_{n}\right)=n$ by the design, and $\sup_{n}\mathrm{{E}}X_{n}=\infty$.

Intuitively, 'bounded in probability' only restricts the probability placed on the extreme values; it does not say anything about the ratio between the extreme value and its associated probability. I hope this example could clarify something for your problem.


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